Maximum Number of Eaten Apples

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Greedy AlgorithmsArrays

There is a special kind of apple tree that grows apples every day for n days. On the i^th day, the tree grows apples[i] apples that will rot after days[i] days, that is on day i + days[i] the apples will be rotten and cannot be eaten. On some days, the apple tree does not grow any apples, which are denoted by apples[i] == 0 and days[i] == 0.

You decided to eat at most one apple a day (to keep the doctors away). Note that you can keep eating after the first n days.

Given two integer arrays days and apples of length n, return the maximum number of apples you can eat.

Example 1:

Input: apples = [1,2,3,5,2], days = [3,2,1,4,2]
Output: 7
Explanation: You can eat 7 apples:
- On the first day, you eat an apple that grew on the first day.
- On the second day, you eat an apple that grew on the second day.
- On the third day, you eat an apple that grew on the second day. After this day, the apples that grew on the third day rot.
- On the fourth to the seventh days, you eat apples that grew on the fourth day.

Example 2:

Input: apples = [3,0,0,0,0,2], days = [3,0,0,0,0,2]
Output: 5
Explanation: You can eat 5 apples:
- On the first to the third day you eat apples that grew on the first day.
- Do nothing on the fouth and fifth days.
- On the sixth and seventh days you eat apples that grew on the sixth day.

Constraints:

n == apples.length == days.length
1 <= n <= 2 * 10⁴
0 <= apples[i], days[i] <= 2 * 10⁴
days[i] = 0 if and only if apples[i] = 0.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the `apples` and `days` arrays? What is the maximum value of an element in the `apples` and `days` arrays?
Can the `apples` or `days` arrays be empty or null?
If I have apples that expire on the same day, does the order in which I eat them matter for determining the maximum number of apples I can eat?
If at any point I have no apples to eat, should I advance the day and continue the simulation, or should I terminate early and return the number of apples eaten so far?
If the `days[i]` value is zero, does that imply there are no days after the current day to eat the corresponding `apples[i]`?

Brute Force Solution

Approach

Imagine you have a basket of apples that ripen on different days, and you want to eat as many as possible. The brute force method involves trying every possible eating schedule to see which one lets you eat the most apples. We will explore every possible combination of eating or skipping apples each day.

Here's how the algorithm would work step-by-step:

Start by looking at the first day and decide whether to eat an apple if there are any available or skip eating an apple on that day.
Next, look at the second day and, again, decide whether to eat an apple or skip it. Repeat this for every possible eating decision from the first day.
Keep going through each day, making all possible 'eat' or 'skip' decisions based on which apples are available to be eaten on that particular day based on when they ripened and if they are still good.
Each time you complete a possible eating schedule (a decision for every day), calculate how many apples were eaten.
After trying every single possible eating schedule, compare the total number of apples eaten for each schedule.
Finally, choose the eating schedule that resulted in eating the most apples. This is your answer.

Code Implementation

def max_eaten_apples_brute_force(apples, days):
    max_apples_eaten = 0

    def solve(current_day, eaten_count, available_apples):
        nonlocal max_apples_eaten

        if current_day == len(apples):
            max_apples_eaten = max(max_apples_eaten, eaten_count)
            return

        # Prune expired apples
        available_apples = [(ripen_day, apple_count) for ripen_day, apple_count in available_apples if ripen_day + days[ripen_day] > current_day]

        # Option 1: Skip eating an apple today
        solve(current_day + 1, eaten_count, available_apples[:])

        # Option 2: Eat an apple if available
        if available_apples:
            # Sort to prioritize apples expiring sooner
            available_apples.sort()
            ripen_day, apple_count = available_apples[0]
            # Decrement the apple count or remove if it's the last one
            if apple_count > 1:
                available_apples[0] = (ripen_day, apple_count - 1)
                solve(current_day + 1, eaten_count + 1, available_apples[:])
            else:
                solve(current_day + 1, eaten_count + 1, available_apples[1:])
    
    # Represents apples as (ripen_day, number_of_apples)
    solve(0, 0, [])
    return max_apples_eaten

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores every possible eating schedule by making a decision on each day whether to eat an apple or skip it. For each of the n days, there are two choices (eat or skip). Therefore, the total number of possible schedules grows exponentially with the number of days. This results in 2 * 2 * ... * 2 (n times) which is 2^n. Thus, the time complexity is O(2^n).

Space Complexity

O(2^N) – The brute force approach explores every possible eating schedule through recursion. For each of the N days, there are two choices: eat or skip. This leads to a binary decision tree of depth N. The recursion stack can grow up to a maximum depth of N, but more importantly, the number of different eating schedules (leaf nodes of this binary tree) that need to be generated and processed grows exponentially as 2^N. Therefore, storing all of these schedules (or potentially tracking visited states to avoid redundancy, though not explicitly mentioned, represents a form of memoization) has a space complexity of O(2^N), where N is the number of days.

Optimal Solution

Approach

To maximize the number of eaten apples, we use a strategy that prioritizes eating apples that will rot soonest. We keep track of available apples and always pick the ones with the nearest expiration date first.

Here's how the algorithm would work step-by-step:

Keep track of all the apples available on any given day, along with their expiration dates.
On each day, first check for any apples that have already expired and remove them from consideration.
Then, from the apples still available, choose the apple that will expire the soonest to eat.
If there are no apples available on a given day, skip that day.
Repeat this process for as long as there are apples to consider or days to process. The final count will be the maximum number of eaten apples.

Code Implementation

import heapq

def maximum_number_eaten_apples(apples, days):
    eaten_apples_count = 0
    available_apples = []
    current_day = 0

    while current_day < len(apples) or available_apples:
        # Add apples that become available today
        if current_day < len(apples) and apples[current_day] > 0:
            heapq.heappush(available_apples, (current_day + days[current_day], apples[current_day]))

        # Remove expired apples
        while available_apples and available_apples[0][0] <= current_day:
            heapq.heappop(available_apples)

        # Eat an apple if available
        if available_apples:
            expiration_date, number_of_apples = heapq.heappop(available_apples)
            eaten_apples_count += 1

            # Put the remaining apples back, if any
            if number_of_apples - 1 > 0:
                heapq.heappush(available_apples, (expiration_date, number_of_apples - 1))

        current_day += 1

    return eaten_apples_count

Big(O) Analysis

Time Complexity

O(n log n) – The primary driver of time complexity is the priority queue (or heap) used to store apples and their expiration dates. In the worst case, we iterate through all n days (where n is the length of the input arrays). On each day, we potentially add a new apple to the priority queue, which takes O(log n) time. We also potentially remove expired apples and extract the apple with the earliest expiration, both operations taking O(log n) time as well. Therefore, the overall time complexity is dominated by the heap operations performed over n days, resulting in O(n log n).

Space Complexity

O(N) – The algorithm keeps track of available apples along with their expiration dates. A priority queue or similar data structure (e.g., a min-heap) would be suitable for selecting the apple expiring soonest. In the worst case, we might have apples available for every day, resulting in a data structure that stores information for up to N apples, where N is the number of days (length of the input arrays). Therefore, the auxiliary space used is proportional to N. This leads to a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty arrays for days and apples	Return 0 immediately as no apples can be eaten without days or apples.
days.length != apples.length	Pad the shorter array with zeros to align the lengths, representing days with no available apples or vice versa.
All days have zero apples	Return 0, as no apples can be eaten if none are available on any day.
apples[i] is very large (integer overflow)	Ensure the data type used to store the number of apples eaten doesn't overflow by using a long/BigInteger if necessary.
days[i] is very large (potential for huge heap)	Use a priority queue of size at most 'largest day' to limit space usage and avoid excessive allocation.
days are not monotonically increasing, representing a time warp	The solution must process the input arrays in their given order, respecting the day each apple becomes available, regardless of day values.
A day comes with an extremely large amount of apples but expires almost immediately	Ensure the priority queue correctly handles apples with very short expiration periods to maximize apple consumption.
All apples expire on the same day	Handle by picking one apple to eat each day to maximize the duration eating apples.

Empty arrays for days and apples

How to Handle:

Return 0 immediately as no apples can be eaten without days or apples.

days.length != apples.length

How to Handle:

Pad the shorter array with zeros to align the lengths, representing days with no available apples or vice versa.

All days have zero apples

How to Handle:

Return 0, as no apples can be eaten if none are available on any day.

apples[i] is very large (integer overflow)

How to Handle:

Ensure the data type used to store the number of apples eaten doesn't overflow by using a long/BigInteger if necessary.

days[i] is very large (potential for huge heap)

How to Handle:

Use a priority queue of size at most 'largest day' to limit space usage and avoid excessive allocation.

days are not monotonically increasing, representing a time warp

How to Handle:

The solution must process the input arrays in their given order, respecting the day each apple becomes available, regardless of day values.

A day comes with an extremely large amount of apples but expires almost immediately

How to Handle:

Ensure the priority queue correctly handles apples with very short expiration periods to maximize apple consumption.

All apples expire on the same day

How to Handle:

Handle by picking one apple to eat each day to maximize the duration eating apples.